Tense Operators on de Morgan Algebras

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Tense operators on De Morgan algebras

ALDO V. FIGALLO and GUSTAVO PELAITAY, Departamento deMatematica, Universidad Nacional del Sur, 8000 Baha Blanca and Instituto deCiencias Basicas, Universidad Nacional de San Juan, 5400 San Juan, Argentina

AbstractThe purpose of this article is to investigate the variety of algebras, which we call tense De Morgan algebras, as a naturalgeneralization of tense algebras developed in Burges (1984, Handbook of Philosophical Logic, vol. II, pp. 89139) (see also,Kowalski (1998, Rep. Math. Logic, 32, 5395)). It is worth mentioning that the latter is a proper subvariety of the rst one,as it is shown in a simple example. Our main interest is the representation theory for these classes of algebras.

Keywords: De Morgan algebras, tense operators, frames, duality theory.

AMS subject classication (2000): 06D30, 06D50, 03B44.

1 Introduction

Propositional logics usually do not incorporate the dimension of time; consequently, in order to obtaina tense logic, we enrich a propositional logic by the addition of new unary operators (or connectives)which are usually denoted byG,H ,F and P. We can dene F and P by means ofG andH as follows:F(x)=G(x) and P(x)=H (x), where x denotes negation of the proposition x.

Tense algebras are algebraic structures corresponding to the propositional tense logic (see [5, 21]).An algebra A,,,,G,H ,0,1 is a tense algebra if A,,,,0,1 is a Boolean algebra and G,H are unary operators on A which satisfy these axioms:

G(1)=1, H (1)=1,G(xy)=G(x)G(y), H (xy)=H (x)H (y),

xGP(x), xHF(x),where P(x)=H (x) and F(x)=G(x).

In the last few years tense operators have been considered by different authors for various classesof algebras. Some contributions in this area have been the papers by Botur et al. [4], Chajda [6],Chirita [7, 8], Diaconescu and Georgescu [10] and Figallo et al. [1216].

In the present article we consider tense operators on De Morgan algebras. De Morgan algebrasare related to logic and have been studied by several authors (see [1, 3, 9, 20, 22]). In particular,they are related with a four-valued logic developed by Belnap in [2].

The article is organized as follows. In Section 1 we summarize the main denitions and resultsneeded in this article. In Section 2 we dene the variety of tense De Morgan algebras and introducesome examples. In Section 3 we study the representation of tense De Morgan algebras. In Section 4we describe a topological duality for tense DeMorgan algebras. Finally, in Section 5, we characterizethe congruences of these algebras in terms of the mentioned duality and certain closed subsets ofthe space associated with them.

E-mail: [email protected]: [email protected]

Vol. 22 No. 2, The Author 2013. Published by Oxford University Press. All rights reserved.For Permissions, please email: [email protected]:10.1093/jigpal/jzt024 Advance Access published 29 July 2013

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256 Tense operators on De Morgan algebras

2 Preliminaries

Given a relational structure X , where X = and is a reexive and transitive binary relationon X (i.e. a quasi-order set), we will denote by (Y ] ([Y )) the set {xX :yY xy} ([Y )={xX :yY yx}), for any Y X .

A set Y X is increasing if Y =[Y ), and it is decreasing if Y = (Y ]. The distributive lattice of allincreasing subsets of X , will be denoted by Pi(X ) and the power set of X by P(X ). If Y is asubset of X , then Y c will denote the set-theoretical complement of Y , i.e. Y c =X \Y . Let R and Qbe binary relations dened on a set X . The composition of R with Q is denoted by RQ. Let X ,Ybe sets. Given a relation RX Y , for each Z X , R(Z) we will denote the image of Z by R. IfZ ={x}, we will write R(x) instead of R({x}). Moreover, for each V Y , R1(V ) we will denote theinverse image of V by R, i.e. R1(V )={xX :R(x)V =}. If V ={y}, we will write R1(y) insteadof R1({y}).

An algebra A,,,,0,1 is a De Morgan algebra if A,,,0,1 is a bounded distributivelattice and is a unary operation on A satisfying the following identities:1. (xy)=xy;2. x=x; and3. 0=1.In what follows a De Morgan algebra A,,,,0,1 will be denoted briey by (A,).A DeMorgan frame (see [11]) is a structure X ,,g, where X , is a quasi-order and g :X X

is a function such that g(g(x))=x, and if xy, then g(y)g(x), for all x,yX . Let A be a DeMorganalgebra and let X (A) be the set of all prime lters of A. It is known that X (A),,gA is a De Morganframe, where gA :X (A)X (A) is the involution dened by

gA(S)={xA :x /S}, for all SX (A). (1)

Moreover, if X ,,g is a De Morgan frame, then Pi(X ),,,g,,X is a De Morgan algebra,where g :Pi(X )Pi(X ) is dened by

g U =X \g(U ), for allU Pi(X ). (2)

Even though the theory of Priestley spaces and its relation to bounded distributive lattices arewell known (see [1719]), we will recall some denitions and results with the purpose of xing thenotations used in this article.

Let us recall that a Priestley space (or P-space) is a compact totally disconnected ordered topo-logical space. If X is a P-space and D(X ) is the family of increasing, closed and open subsets of aP-space X , then D(X ),,,,X is a bounded distributive lattice.

On the other hand, let A be a bounded distributive lattice and X (A) the set of all prime ltersof A. Then X (A), ordered by set inclusion and with the topology which has as a sub-basis thesets A(a)={SX (A) :aS} and X (A)\A(a), for each aA, is a P-space. In addition the mappingA :AD(X (A)) is a lattice isomorphism. Besides, if X is a P-space, the mapping X :X X (D(X ))dened by X (x)={U D(X ) :xU }, is a homeomorphism and an order isomorphism.

If we denote by L the category of bounded distributive lattices and their corresponding homomor-phisms and by P the category of P-spaces and the continuous increasing mappings (or P-functions),then there exists a duality between both categories by dening the contravariant functors :P L

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Tense operators on De Morgan algebras 257

and :LP as follows:(P1) For each P-space X , (X )=D(X ), and for every P-function f :X1X2, (f )(U )= f 1(U )

for all U D(X2).(P2) For each bounded distributive lattice A, (A)=X (A), and for every bounded lattice homomor-

phism h :A1A2, (h)(S)=h1(S) for all SX (A2).On the other hand, Priestley proved that, if A is a bounded distributive lattice and Y is a closed

subset of X (A), then

(P3) (Y )={(a,b)AA :A(a)Y =A(b)Y } is a congruence on A and the correspondenceY (Y ) is an anti-isomorphism from the lattice of all closed sets of X (A) onto the lattice ofall congruences on A.

In 1977, Cornish and Fowler [9] extended the Priestley duality to De Morgan algebras consideringDe Morgan spaces (or mP-spaces) as pairs (X ,g), where X is a P-space and g is an involutivehomeomorphism of X and an anti-isomorphism. They also dened the mP-functions from an mP-space (X1,g1) into another, (X2,g2), as continuous and increasing functions (P-functions) f from X1into X2, which satisfy the additional condition f g1=g2f .

In addition, these authors introduced the notion of an involutive set in an mP-space (X ,g) as asubset Y of X such that Y =g(Y ) and characterized the congruences of a De Morgan algebra (A,)by means of the family CI (X (A)) of involutive closed subsets of X (A). To achieve this result, theyproved that the function I from CI (X (A)) onto the family ConM (A) of congruences on A denedas in (P3) is a lattice anti-isomorphism.

3 Tense De Morgan algebras

Let (A,) be a De Morgan algebra and G,H :AA two unary operators. We dene the operatorsF and P by F(x)=G(x) and P(x)=H (x), for any xA.DEFINITION 3.1An algebra (A,,G,H ) is a tense De Morgan algebra if (A,) is a De Morgan algebra and G andH are two unary operations on A such that:

(T1) G(1)=1 and H (1)=1;(T2) G(xy)=G(x)G(y) and H (xy)=H (x)H (y);(T3) xGP(x) and xHF(x); and(T4) G(xy)G(x)F(y) and H (xy)H (x)P(y).

We will denote these algebras by (A,,G,H ) or simply by A if no confusion may arise.EXAMPLE 3.2It is easy to see that every tense algebra is also a tense De Morgan algebra.

EXAMPLE 3.3There are two extreme examples of tense operators on a De Morgan algebra (A,). Dene G=H ,such that G(1)=1 and G(x)=0 for x =1. We can check that (A,,G,H ) is a tense De Morganalgebra. Another example of tense operators are identical mappings, i.e. G(x)=x=H (x) for allxA.We will indicate an example of tense De Morgan algebra which is not a tense algebra.

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EXAMPLE 3.4Let us consider the De Morgan algebra ({0,a,b,c,d,1},), which is described as follows:

.................................................................................

....................................................................................................................................................................................................................................................................................................................................

......................

......................

......................

...............

.................................................................................

a

c

b

d

0

1

x x0 1a db cc bd a1 0

Dene operators G, H by the table

x G(x) H(x)0 a 0a a 0b a bc c bd a 11 1 1

Then, ({0,a,b,c,d,1},,G,H ) is a tense De Morgan algebra.

PROPOSITION 3.5The following properties hold in any tense De Morgan algebra (A,,G,H ) :(T5) xy implies G(x)G(y) and H (x)H (y);(T6) xy implies P(x)P(y) and F(x)F(y);(T7) F(0)=0 and P(0)=0;(T8) F(xy)=F(x)F(y) and P(xy)=P(x)P(y);(T9) FH (x)x and PG(x)x; and

(T10) G(x)F(y)F(xy) and H (x)P(y)P(xy).PROOF. It is routine.

4 Representation by sets

In this section we will show a representation theorem for tense De Morgan algebras in terms of tenseDe Morgan algebras of sets, using the well-known representation theorem for De Morgan algebras.

Let X be a set, R be a binary relation on X and R1 the converse of R. We dene four operatorson P(X ) as follows:

GR(U )={xX :R(x)U }, HR1 (U )={xX :R1(x)U }.FR(U )={xX :R(x)U =}, PR1 (U )={xX :R1(x)U =}.

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In general, the De Morgan algebra (Pi(X ),g) is not closed under the operators introduced above.In the next result we will give conditions on the relation R and its converse for that Pi(X ) be closedunder these operations.

PROPOSITION 4.1Let X , be a poset, R a binary relation on X , and R1 the converse of R. Then(i) (R) (R) if and only if GR(U )Pi(X ), for all U Pi(X ).(ii) (R1) (R1) if and only if HR1 (U )Pi(X ), for all U Pi(X ).(iii) (1 R) (R1) if and only if FR(U )Pi(X ), for all U Pi(X ).(iv) (1 R1) (R11) if and only if PR1 (U )Pi(X ), for all U Pi(X ).PROOF. We will only prove (i). Let us show that, if U Pi(X ), then GR(U )Pi(X ). Let xy andR(x)U . Let zX , such that (y,z)R. Then, there exists wX such that (x,w)R and wz.As R(x)U , wU ; since U is increasing, zU . Conversely, suppose that GR(U )Pi(X ), for allU Pi(X ). Let x,y,zX such that xy and (y,z)R. Let us consider the increasing set [R(x)). AsR(x)[R(x)), we have that xGR([R(x))). Then, yGR([R(x))). Thus, z[R(x)), i.e. there existswX such that (x,w)R and wz, i.e. (x,z) (R).DEFINITION 4.2A structure F =X ,,g,R,R1 is a tense De Morgan frame if X ,,g is a De Morgan frame, Ris a binary relation on X , and R1 is the converse of R such that:

(F1) (R) (R);(F2) (R1) (R1); and(F3) (x,y)R implies (g(x),g(y))R.REMARK 4.3In every tense De Morgan frame F =X ,,g,R,R1 the following conditions are satised:(F4) (1 R) (R1); and(F5) (1 R1) (R11).PROPOSITION 4.4Let F =X ,,g,R,R1 be a tense De Morgan frame. Then,(i) for all U Pi(X ), FR(U )=g GR(g U ); and(ii) for all U Pi(X ), PR1 (U )=g HR1 (g U ).PROOF. We will prove only (i). Let xFR(U ) and suppose that x /g GR(g U ), i.e. R(g(x))gU . Then, there is zU such that (x,z)R. By (F3) we have that (g(x),g(z))R. Then, we candeduce that g(z)g U , which is a contradiction. Therefore, xg GR(g U ). Conversely, supposethat xg GR(g U ). Thus, g(x) /GR(g U ), i.e. R(g(x)) g U . From the last assertion, there iszR(g(x)) such that g(z)U . Thus, by (F3), we deduce that g(z)R(x)U . Therefore, xFR(U ).

PROPOSITION 4.5If F =X ,,g,R,R1 is a tense De Morgan frame, then A(F)= (Pi(X ),g,GR,HR1 ) is a tense DeMorgan algebra.

PROOF. Taking into account Proposition 4.1, Remark 4.3 and Proposition 4.4, we only have toprove (T4). Indeed: let xGR(U V ) and suppose that x /GR(U ). Then, there exists yX such that(x,y)R and y /U . Since, yU V , we have that yV . Therefore, xFR(V ). In a similar way wecan prove HR1 (U V )HR1 (U )PR1 (V ).

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DEFINITION 4.6Let (A,,G,H ) be a tense De Morgan algebra. We will dene two binary relations RAG and RAH onX (A) as follows:

(S,T )RAG if and only if G1(S)T F1(S),(S,T )RAH if and only if H1(S)T P1(S).

The following Proposition is fundamental for the proof to Proposition 4.9.

PROPOSITION 4.7Let (A,,G,H ) be a tense De Morgan algebra. Then the following conditions are equivalents:(i) (S,T )RAG; and(ii) (T ,S)RAH .PROOF. (i) (ii). Let S,T be two prime lters of A such that G1(S)T F1(S) and let us supposethat H (x)T . Thus, FH (x)S. From the preceding assertion and (T9), we have that xS. Then,H1(T )S. On the other hand, suppose that zS. By (T3), GP(z)S and, since G1(S)T , wehave that P(z)T . Therefore, SP1(T ). The converse implication is similar.REMARK 4.8From Proposition 4.7 we have that RAH is the converse of R

AG .

PROPOSITION 4.9Let (A,,G,H ) be a tense De Morgan algebra. Then, F(A)=X (A),,gA,RAG,RAH is a tense DeMorgan frame.

PROOF. Taking into account Remark 4.8, we only have to prove (F1), (F2) and (F3).(F1): Let S,T ,D be three prime lters of A such that SD andG1(D)T F1(D). Let us considerthe ideal

(TcF1(S)c]

and prove that

G1(S)(TcF1(S)c]=. (3)

Suppose that (3) is not fullled. Then there exist elements xG1(S), yTc and z /F1(S) suchthat xyz. So, by (T5) and (T4), we have that G(x)G(yz)G(y)F(z). Since G(x)S andF(z) /S, we deduce that G(y)S, which is impossible. Then by the BirkhoffStone theorem thereis a prime lter Z such that

G1(S)Z and (TcF1(S)c)Z =

Therefore, G1(S)Z F1(S) andZ T ., i.e. (S,T ) (RAG).(F2): It is proved in a similar way to (F1).

(F3): Let S,T be two prime lters such that G1(S)T F1(S). Suppose that xG1(gA(S)). So,G(x)=F(x) /S, and since T F1(S), we have that x /T , i.e. xgA(T ). On the other hand,suppose that zgA(T ) and F(z) /gA(S). So, F(z)=G(z)S, and since G1(S)T , we have that

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zT , i.e. z /gA(T ), which is a contradiction. Thus, F(z)gA(S). Therefore, G1(gA(S))gA(T )F1(gA(S)).

The following result is necessary for the proof to Theorem 4.11.

LEMMA 4.10Let (A,,G,H ) be a tense De Morgan algebra and let SX (A) and aA. Then,(i) G(a) /S if and only if there exists T X (A) such that (S,T )RAG and a /T ; and(ii) H (a) /S if and only if there exists Z X (A) such that (S,Z)RAH and a /Z .PROOF. Suppose that G(a) /S. Let us consider the ideal ({a}F1(S)c], and we will prove that

G1(S)({a}F1(S)c]=. (4)Suppose the opposite. Then there exist elements bG1(S) and cF1(S)c such that bac.From (T5) and (T4) we have that

G(b)G(ac)G(a)F(c).From this assertion we deduce that F(c)S, which is a contradiction. Thus (4) is veried.

Therefore from the BirkhoffStone theorem there is a prime lter T such that

G1(S)T and ({a}F1(S)c)T =.Then, (S,T )RAG and a /T . The other implication is easy. In a similar way we can prove (ii).

Let (A,,G,H ) be a tense De Morgan algebra and let us consider its associated frame F(A)=X (A),,gA,RAG,RAH . The structure

A(F(A))= (Pi(X (A)),gA ,GRAG ,HRAH )is a tense De Morgan algebra.

Let A be a De Morgan algebra. For each aA consider the setA(a)={SX (A) :aS}

and the family of sets

(A)={A(a) :aA}.It is known that the structure

((A),gA )is a De Morgan subalgebra of the De Morgan algebra

Pi(X (A))= (Pi(X (A)),gA ).The assignment A :APi(X (A)) is an injective homomorphism of De Morgan algebras whose

range is the set (A). Moreover, this result can be extended to tense De Morgan algebras as follows.

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THEOREM 4.11For every tense De Morgan algebra (A,,G,H ) the algebra

(A)= ((A),gA ,GRAG ,HRAH )

is a subalgebra of A(F(A)) isomorphic to A by means of the assignment A :A(A).PROOF. It is easy to check that (A) is a subalgebra of A(F(A)) and that the mapping A is injective.We will prove that A is a homomorphism of tense De Morgan algebras. We need only to see thatA(G(a))=GRAG (A(a)) and A(H (a))=HRAH (A(a)). Now we will prove the inclusion GRAG (A(a))A(G(a)). Let us take a prime lter S such that G(a) /S. From Lemma 4.10, there exists T X (A) such that (S,T )RAG and a /T . Then, we have that RAG(S) A(a). Thus S /GRAG (A(a)). Theinclusion A(G(a))GRAG (A(a)) is immediate. The equality A(H (a))=HRAH (A(a)) follows by asimilar argument using (ii) of Lemma 4.10.

5 Duality for tense De Morgan algebras

In this section, we will indicate a duality for tense De Morgan algebras taking into account theresults established by Cornish and Fowler in [9].

DEFINITION 5.1A tense De Morgan space (or tmP-space) is a system (X ,g,R,R1) where (X ,g) is an mP-space, Ris a binary relation on X and R1 the converse of R such that:

(S1) For each U D(X ) it holds that GR(U ), HR1 (U )D(X ), where GR(U ), HR1 (U ) are denedas in Section 3.

(S2) (x,y)R implies (g(x),g(y))R.(S3) For each xX , R(x) is a closed set.(S4) For each xX , R(x)= (R(x)][R(x)).DEFINITION 5.2A tmP-function from a tmP-space (X1,g1,R1,R

11 ) into another one, (X2,g2,R2,R

12 ), is a mP-function

f :X1X2 which satises the following conditions:(r1) (x,y)R1 implies (f (x),f (y))R2 for x,yX1;(r2) if (f (x),y)R2, then there is an element zX such that (x,z)R1 and f (z)y; and(r3) if (y,f (x))R2, then there is an element zX1 such that (z,x)R1 and f (z)y.

Next we will show that the category tmPS of tmP-spaces and tmP-functions is dually equivalentto the category TDMA of tense De Morgan algebras and tense De Morgan homomorphisms.

LEMMA 5.3Let (X ,g,R,R1) be a tmP-space. Then, (D(X ),g,GR,HR1 ) is a tense De Morgan algebra, wherefor all U D(X ), g U is the set dened in (2).PROOF. The denition of GR and HR1 are consequences of (S1). Conditions (T3) and (T4) give usthe equation (S2). Finally, (T1) and (T2) are consequences of the denitions of GR and HR1 .

LEMMA 5.4Let f : (X1,g1,R1,R11 ) (X2,g2,R2,R12 ) be a morphism of tmP-spaces. Then (f ) :D(X2)D(X1)dened as in (P1) is a tense De Morgan morphism.

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PROOF. We only prove that f 1(GR2 (U ))=GR1 (f 1(U )). The inclusion f 1(GR2 (U ))GR1 (f 1(U ))follows by (r1). Let xGR1 (f 1(U )) and let yX2 such that (f (x),y)R2. By (r2), there existszX1 such that (x,z)R1 and f (z)y. Since zR1(x) f 1(U ), yU . Thus f (x)GR2 (U ), andconsequently GR1 (f

1(U )) f 1(GR2 (U )).

The previous two lemmas show that is a contravariant functor from TDMA to tmPS.Let RD(X ) be the relation dened in X (D(X )) by means of the operator GR, i.e.

(X (x),X (y))RD(X )G1R (X (x))X (y)F1R (X (x)).PROPOSITION 5.5Let (X ,g,R,R1) be a tmP-space. Then the following conditions are equivalent:

(i) for all xX , R(x) is closed and R(x)= (R(x)][R(x)).(ii) for all xX , R(x) is compact and R(x)= (R(x)][R(x)).(iii) for every x,yX , if (X (x),X (y))RD(X ) then (x,y)R.PROOF. (i) (ii) It is immediate. (ii) (iii). Suppose that (X (x),X (y))RD(X ) and y /R(x)= (R(x)][R(x)). If y /[R(x)), then for all zR(x), z y. As X is a Priestley space, for each zR(x) there existsUz D(X ) such that zUz and y /Uz. Then, R(x)

xR(x)Uz and y /

zR(x)Uz. Since R(x) is compact,

R(x)U for some U D(X ) such that y /U . But as (X (x),X (y))RD(X ) and xGR(U ), then yU ,which is impossible. If y / (R(x)], by a similar argument, we obtain also a contradiction. (iii) (i).Weprove that (R(x)][R(x))R(x); the other inclusion always holds. Suppose that y (R(x)][R(x)).Then there exists z1,z2X such that

z1yz2 and z1,z2R(x).

By assumption, (X (x),X (z1))RD(X ), (X (x),X (z2))RD(X ) and also X (z1)X (y)X (z2). It iseasy to check that (X (x),X (y))RD(X ). It follows from the assumption (iii) that (x,y)R. ThereforeyR(x). We now prove that R(x) is closed. Suppose that yCl(R(x)) and y /R(x), where Cl(R(x))denote the closure of the set R(x). By the asumption (iii), we have that (X (x),X (y)) /RD(X ). So,there exists U D(X ) such that GR(U )X (x) and U /X (y), or there exists V D(X ) such thatV X (y) and FR(V ) /X (x). In the rst case R(x)U and yUc. It follows that R(x)Uc = andyUc, which is impossible, because yCl(R(x)). In the other case, yV and R(x)V =, whichis a contradiction. Therefore, R(x) is a closed subset of X .

LEMMA 5.6Let (A,,G,H ) be a tense De Morgan algebra. Then (X (A),,gA,RAG,RAH ) is a tmP-space, wheregA, RAG and R

AH are dened as in (1) and Denition 4.6, respectively.

PROOF. We will only prove (S3). Let us check that RAG(S) is closed for SX (A). Suppose thatT /RAG(S). Then by denition of the relation RAG , there exists xG1(S) such that x /T or there existsyT such that y /F1(S). In the rst case RAG(S)A(x) and in the second case RAG(S)X (A)\A(y).

LEMMA 5.7Let h :A1A2 be a tense De Morgan morphism. Then, (h) :X (A2)X (A1) dened as in (P2) isa morphism of tmP-spaces.

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PROOF. We will prove (r1) and (r2); (r3) can be proved in a similar way to (r2). Let S,T X (A2).Let us prove that if (S,T )RA2G , then (h1(S),h1(T ))RA1G . Suppose that aG1(h1(S)). Then wehave that h(G(a))=G(h(a))S, from which it follows that h(a)T , i.e. ah1(T ). In a similar waywe can prove h1(T )F1(h1(S)). Now suppose that (h1(S),T )RA1G . Let us prove that thereexists Z X (A2) such that (S,Z)RA2G and h1(Z)T . First we will prove the existence of the lterZ . The proof will be consists of two parts:

1. We will prove the existence of a prime lter DA2, such that G1(S)D and h1(D)T .2. We will prove the existence of a prime lter Z A2, such that G1(S)Z F1(S) and Z D.From 1 and 2, the existence of the prime lter Z follows, which fullls the condition will be

established.By hypothesis we have that G1(h1(S))T F1(h1(S)). Now let us prove that

G1(S)(h(Tc)]=. (5)Suppose that (5) not satised. Consequently there exist elements aG1(S), bTc, such that

ah(b). Then, G(a)G(h(a))S. From this it follows that, bG1(h1(S))T , which is a con-tradiction. Then (5) holds. By the BirkhoffStone theorem, there is a prime lter D, such that

G1(S)D and h1(D)T .Now we will prove the second part. For this purpose let us rst prove that

G1(S)(DcF1(S)c]=. (6)Suppose the opposite. Then, there exist elements aG1(S), bDc and cF1(S)c such that

abc.Then, we have

G(a)G(bc)G(b)F(c)S.

Since b /D and G1(S)D, G(b) /S. Therefore, F(c)S, which is a contradiction. Then, (6)holds. By applying the BirkhoffStone theorem, we can make sure that there is a prime lter Z suchthat

G1(S)Z F1(S) andZ D.

As h1(Z)h1(D)T , we have proved the existence of the prime lter Z .

The two previous lemmas show that is a contravariant functor from tmPS to TDMA.Now, we will see that these categories are dually equivalent to each other.

PROPOSITION 5.8Let (X ,g,R,R1) be a tmP-space; then, X :X X (D(X )) is an isomorphism of tmP-spaces.PROOF. It follows from the results established in [9] and Proposition 5.5.

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PROPOSITION 5.9Let (A,,G,H ) be a tense De Morgan algebra; then, A :AD(X (A)) is a tense De Morgan iso-morphism.

PROOF. It follows from the results established in [9] and Lemma 4.10.

The following theorem follows from Propositions 5.8 and 5.9.

THEOREM 5.10Functors X and A establish a dual equivalence between the categories TDMA and tmPS.

6 Application of the duality

In this section we will use the duality described in Section 4 to characterize the congruence latticeCon(A) of a tense De Morgan algebra A.

DEFINITION 6.1Let (X ,g,R,R1) be a tmP-space. An involutive closed subset Y of X is a tmP-subset of X if itsatises the following conditions for u,vX :(ts1) if (v,u)R and uY , then there exists, wY such that (w,u)R and wv.(ts2) if (u,v)R and uY , then there exists, zY such that (u,z)R and zv.LEMMA 6.2Let (A,,G,H ) be a tense De Morgan algebra and Y , a tmP-subset of X (A). Then (Y ) is a tenseDe Morgan congruence, where (Y ) is dened as in (P3).

PROOF. We will only prove that (Y ) preserves G and H . Let (a,b)(Y ) and suppose that SGRAG (A(a))Y . Hence, (RAG)1(S)A(a) and SY . Suppose that T (RAG)1(S). Consequently,from (ts1) there is W Y , such that W T and W (RAG)1(S). This last assertion allows us toinfer that W A(a), from which we conclude W A(b)Y . Therefore, since W T , we have thatT A(b); hence, SGRAG (A(b))Y . Then, GRAG (A(a))Y GRAG (A(b))Y . The other inclusion isproved in a similar way. Analogously, (Y ) preserves H .

LEMMA 6.3Let (A,,G,H ) be a tense De Morgan algebra and Con(A). If q :AA/ is the natural epimor-phism, then Y ={(q)(S) :SX (A/ )} is a tmP-subset, where is dened as in (P2).PROOF. Since ConM (A) is a sublattice of Con(A) we have that Y ={(q)(S) :SX (A/ )} is aninvolutive closed subset of X (A) and =(Y ) (see [9]). Besides, from Lemma 5.7 we have that(q) is a tmP-function. In addition, Y is a tmP-subset of X (A). Indeed, let (V ,U )RAG and U Y .From the last assertion, there is SX (A/ ) such that (q)(S)=U . Then, (V ,(q)(S))RAG . As aconsequence, we have from (r2) that there is T X (A/ ) such that (T ,S)RA/G and (q)(T )V .Therefore, we have from (r1) that (q)(T ) (RAG)1(U ). (ts2) can be proved in a similar way.

From Lemmas 6.2 and 6.3, we obtain Theorem 6.4.

THEOREM 6.4Let (A,,G,H ) be a tense De Morgan algebra and (X (A),gA,RAG,RAH ), the associated tmP-space ofA. Then, there is an anti-isomorphisms between Con(A) and the lattice of all tmP-subset of X (A).

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Acknowledgement

The support of CONICET is gratefully acknowledged by Gustavo Pelaitay.

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[20] J. Kalman. Lattices with involution. Transactions of the American Mathematical Society, 87,485491, 1958.

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Received 31 January 2013

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1 Introduction2 Preliminaries3 Tense De Morgan algebras4 Representation by sets5 Duality for tense De Morgan algebras6 Application of the duality

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Tense Operators on de Morgan Algebras